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(D) The equation of the line with intercepts $a$ and $b$ is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $\frac{1}{a}x + \frac{1}{b}y - 1 =

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$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}$

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(D) The equation of the line with intercepts $a$ and $b$ is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $\frac{1}{a}x + \frac{1}{b}y - 1 = 0$.The length of the perpendicular $p$ from the origin $(0, 0)$ to this line is given by $p = \frac{|\frac{1}{a}(0) + \frac{1}{b}(0) - 1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}}$.This simplifies to $p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.Squaring both sides,we get $p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}}$.Therefore,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}$.

If the length of the perpendicular drawn from the origin to the line whose intercepts on the axes are $a$ and $b$ is $p$,then

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